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Q.
The value of the determinant $\begin{vmatrix}a&a+2b&a+4b\\ a+2b&a+4b&a+6b\\ a+4b&a+6b&a+8b\end{vmatrix}$ is equal to
Determinants
Solution:
Operating $R_{2} \to R_{2} -R_{1} $
and $R_{3 } \to R_{3} - R_{2} ,$ we get
$\begin{vmatrix}a&a+2b&a+4b\\ 2b&2b&2b\\ 2b&2b&2b\end{vmatrix}$
which is certainly 0 as $R_2 = R_3$.